June 2018 Paper 1 Q13
13 Solve the simultaneous equations
\[xy = 2 \quad \text{and} \quad y = 3x + 5\]Do not use trial and improvement.
You must show your working. [6 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(3x + 5 = \dfrac{2}{x}\) or \(x(3x + 5) = 2\) | M1 | oe |
| \(3x^2 + 5x - 2\ (= 0)\) or \(3x^2 + 5x = 2\) | M1dep | |
| \((3x + a)(x + b)\ (= 0)\) | M1dep | \(ab = -2\) or \(a + 3b = 5\) |
| \((3x - 1)(x + 2)\ (= 0)\) | A1 | |
| \(x = \tfrac{1}{3} \quad x = -2\) or \(x = \tfrac{1}{3} \quad y = 6\) or \(x = -2 \quad y = -1\) | A1 | |
| \(x = \tfrac{1}{3} \quad x = -2\) and \(y = 6 \quad y = -1\) or \(x = \tfrac{1}{3} \quad y = 6\) and \(x = -2 \quad y = -1\) | A1 | either correct \(x\)’s and correct \(y\)’s or correct coordinate pairs |
| Alternative method 2 | ||
| \(3x + 5 = \dfrac{2}{x}\) or \(x(3x + 5) = 2\) | M1 | oe |
| \(3x^2 + 5x - 2\ (= 0)\) or \(3x^2 + 5x = 2\) | M1dep | |
| \(x = \dfrac{-5 \pm \sqrt{[(5)^2 - 4(3)(-2)]}}{2(3)}\) | M1dep | allow one sign error … but the \(2 \times 3\) term must be beneath the full numerator |
| \(x = \dfrac{-5 \pm 7}{6}\) | A1 | |
| \(x = \tfrac{1}{3} \quad x = -2\) or \(x = \tfrac{1}{3} \quad y = 6\) or \(x = -2 \quad y = -1\) | A1 | |
| \(x = \tfrac{1}{3} \quad x = -2\) and \(y = 6 \quad y = -1\) or \(x = \tfrac{1}{3} \quad y = 6\) and \(x = -2 \quad y = -1\) | A1 | either correct \(x\)’s and correct \(y\)’s or correct coordinate pairs |
| Alternative method 3 | ||
| \(3x + 5 = \dfrac{2}{x}\) or \(x(3x + 5) = 2\) | M1 | oe |
| \(3x^2 + 5x - 2\ (= 0)\) or \(3x^2 + 5x = 2\) | M1dep | |
| \((3 \times)\ (x + {}^{5}\!/_{6})^2 \ldots\ldots\) | M1dep | |
| \(x + {}^{5}\!/_{6} = \pm\,{}^{7}\!/_{6}\) | A1 | |
| \(x = \tfrac{1}{3} \quad x = -2\) or \(x = \tfrac{1}{3} \quad y = 6\) or \(x = -2 \quad y = -1\) | A1 | |
| \(x = \tfrac{1}{3} \quad x = -2\) and \(y = 6 \quad y = -1\) or \(x = \tfrac{1}{3} \quad y = 6\) and \(x = -2 \quad y = -1\) | A1 | either correct \(x\)’s and correct \(y\)’s or correct coordinate pairs |
| Alternative method 4 | ||
| \(y = 3\left(\dfrac{2}{y}\right) + 5\) or \(\dfrac{y(y - 5)}{3} = 2\) | M1 | oe |
| \(y^2 - 5y - 6 = 0\) or \(y^2 - 5y = 6\) | M1dep | |
| \((y + a)(y + b)\ (= 0)\) | M1dep | \(ab = -6\) or \(a + b = -5\) |
| \((y - 6)(y + 1)\ (= 0)\) | A1 | |
| \(y = 6 \quad y = -1\) or \(y = 6 \quad x = \tfrac{1}{3}\) or \(y = -1 \quad x = -2\) | A1 | |
| \(x = \tfrac{1}{3} \quad x = -2\) and \(y = 6 \quad y = -1\) or \(x = \tfrac{1}{3} \quad y = 6\) and \(x = -2 \quad y = -1\) | A1 | either correct \(x\)’s and correct \(y\)’s or correct coordinate pairs |
| Alternative method 5 | ||
| \(y = 3\left(\dfrac{2}{y}\right) + 5\) or \(\dfrac{y(y - 5)}{3} = 2\) | M1 | oe |
| \(y^2 - 5y - 6 = 0\) or \(y^2 - 5y = 6\) | M1dep | |
| \(y = \dfrac{5 \pm \sqrt{[(-5)^2 - 4(1)(-6)]}}{2(1)}\) | M1dep | allow one sign error … but the \(2 \times 1\) term must be beneath the full numerator |
| \(y = \dfrac{5 \pm 7}{2}\) | A1 | |
| \(y = 6 \quad y = -1\) or \(y = 6 \quad x = \tfrac{1}{3}\) or \(y = -1 \quad x = -2\) | A1 | |
| \(x = \tfrac{1}{3} \quad x = -2\) and \(y = 6 \quad y = -1\) or \(x = \tfrac{1}{3} \quad y = 6\) and \(x = -2 \quad y = -1\) | A1 | either correct \(x\)’s and correct \(y\)’s or correct coordinate pairs |
| Alternative method 6 | ||
| \(y = 3\left(\dfrac{2}{y}\right) + 5\) or \(\dfrac{y(y - 5)}{3} = 2\) | M1 | oe |
| \(y^2 - 5y - 6 = 0\) or \(y^2 - 5y = 6\) | M1dep | |
| \((y - {}^{5}\!/_{2})^2 \ldots\ldots\) | M1dep | |
| \(y - {}^{5}\!/_{2} = \pm\,{}^{7}\!/_{2}\) | A1 | |
| \(y = 6 \quad y = -1\) or \(y = 6 \quad x = \tfrac{1}{3}\) or \(y = -1 \quad x = -2\) | A1 | |
| \(x = \tfrac{1}{3} \quad x = -2\) and \(y = 6 \quad y = -1\) or \(x = \tfrac{1}{3} \quad y = 6\) and \(x = -2 \quad y = -1\) | A1 | either correct \(x\)’s and correct \(y\)’s or correct coordinate pairs |
Additional guidance
| Trial and improvement … 0 marks No working shown ….. 0 marks The instructions were clearly stated in the question. |